We have a project where our electrician installed low-voltage wire on the rough-in for two miniature LED lights. He installed one LV wire from the future driver location to the first light and then from the first to the second light. Fast forward to the finish. The driver is capable of powering up to three lights, having terminal blocks for three individual wires from three lights. After taking apart the driver, we discovered there are actually three individual drivers inside - one for each light. Which would be fine if we would have ran a separate wire to each light. So my question is, is it possible to wire two of these individual drivers together in parallel to power up two lights from one wire? The same 120v switchleg would connect to the inputs on both drivers and both (-) would tie together along with both (+) being tied together on the output side. Called WAC Lighting(it's their product), and they said it would not work, but didn't give an explanation why. It seems, in theory, it should work - output voltage would remain the same but output capacity would double. Hoping somebody smarter than me can weigh in. Thanks!
Paralleling LED Drivers?
- Thread starter SPierce
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LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
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- Electrical Contractor
LED drivers are "smarter" than basic power supplies, and they may fight each other trying to maintain the output voltage and current, unlike simply paralleling batteries. You could always try it. Another option is to use a single driver with greater capacity.
gar
Senior Member
- Location
- Ann Arbor, Michigan
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- EE
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SPierce:
At this point you do not know what the I-V curve for this LED bulb is.
That it has such a low voltage rating it may not be good to parallel bulbs. Each bulb may need to be driven by an approximately constant current source. That is sort of implied by the voltage information on your driver, and that a separate driver is provided for each bulb.
Parallel two bulbs. Assume a bulb can tolerate 0.1 A. That would be 0.38 W. Get a 10 ohm resistor of adequate wattage. At 12 V input that is 12 W, and 1 A. A 1500 W space heater is about this resistance. Connect the two paralled bulbs in series with the 10 ohm resistor. Next you need a variable voltage source from about 0 to 12 V DC. If nothing else use 1.5 V D cells. From D cells we can get about 4.5, 6, 7.5, and 9 V. This should be more than enough range.
Apply 4.5 V the current will be about (4.5-3.8)/10 = 0.07 A. Are the bulbs of equal brightness, likely not. Go to 6 V input. Now current is about 0.22 A. Is the brightness disparity greater? Probably. If the brightness is not too great go to 7.5 V where current is about 0.37 A. My guess not equally distributed between the two bulbs. Try the experiment with a different pair of bulbs.
If the bulbs need to be driven individually from a current source, then you will have seen intensity differences.
A different experiment puts the bulbs in series. Now intensities should be close to equal as current changes. Now you might want to use a 20 ohm resistor and go to 12 V as a maximum. At 12 V and 20 ohms I would be about (12-7.6)/20 = 0.22 A thru each bulb.
The results of these experiments should tell you what you can do.
.
SPierce:
At this point you do not know what the I-V curve for this LED bulb is.
That it has such a low voltage rating it may not be good to parallel bulbs. Each bulb may need to be driven by an approximately constant current source. That is sort of implied by the voltage information on your driver, and that a separate driver is provided for each bulb.
Parallel two bulbs. Assume a bulb can tolerate 0.1 A. That would be 0.38 W. Get a 10 ohm resistor of adequate wattage. At 12 V input that is 12 W, and 1 A. A 1500 W space heater is about this resistance. Connect the two paralled bulbs in series with the 10 ohm resistor. Next you need a variable voltage source from about 0 to 12 V DC. If nothing else use 1.5 V D cells. From D cells we can get about 4.5, 6, 7.5, and 9 V. This should be more than enough range.
Apply 4.5 V the current will be about (4.5-3.8)/10 = 0.07 A. Are the bulbs of equal brightness, likely not. Go to 6 V input. Now current is about 0.22 A. Is the brightness disparity greater? Probably. If the brightness is not too great go to 7.5 V where current is about 0.37 A. My guess not equally distributed between the two bulbs. Try the experiment with a different pair of bulbs.
If the bulbs need to be driven individually from a current source, then you will have seen intensity differences.
A different experiment puts the bulbs in series. Now intensities should be close to equal as current changes. Now you might want to use a 20 ohm resistor and go to 12 V as a maximum. At 12 V and 20 ohms I would be about (12-7.6)/20 = 0.22 A thru each bulb.
The results of these experiments should tell you what you can do.
.
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